The spectrum of He is expected to be similar to that of

H
Li⁺
Na
He⁺

Solution

Both He and Li⁺ contain 2 electrons each therefore their spectrum will be similar.

Which of the following pairs is correctly matched ?

Isoelectronics N^3-, O^2-, Cr^3-
Isotones ₁₄Si³⁰, ₁₅P³¹, ₁₆S³²
Isotopes ₂₀Ca⁴⁰, ₁₉K⁴⁰
Isobars ₈O¹⁶, ₈O¹⁷, ₈O¹⁸

Solution

No. of electrons in N^3-, O^2-and Cr^3- are respectively 10, 10 and 27 so these species are not isoelectronic with each other.

Which one is correct for the de-Broglie equation?

E = λ/p
p = h/λ
E= h/p
none of these

Solution

E = mc² (Einstien relation)....... (1)

E = hv....... (2)

From (1) and (2),
mc²= hv

\(\Rightarrow mc^{2}=\frac{hc}{\lambda }\left (∵ v=\frac{c}{\lambda } \right )\)

⇒m c λ= h

\(\Rightarrow \lambda =\frac{h}{mc}\)

⇒λ=^h⁄_p (∵mc=p)

⇒p=^h⁄_λ

According to Bohr’s theory the energy required for an electron in the Li²⁺ ion to be emitted from n = 2 state is (given that the ground state ionization energy of hydrogen atom is 13.6 eV)

61.2 eV
13.6 eV
30.6 eV
10.2 eV

Solution

Energy of electron in 2nd orbit of Li⁺²=\(-13.6\frac{z^{2}}{n^{2}}\)

\(=\frac{-13.6\times (3)^{2}}{(2)^{2}}= –30.6 eV\)

Energy required= 0 – (–30.6)= 30.6 eV

If the mass number of an element is W and its atomic number is N, then

Number of e^-1= W– N
Number of ₁H¹= W– N
Number of ₀n¹= W– N
Number of ₀n¹= N

Solution

No. of neutrons = Mass number – Atomic number= W– N.

The increasing order for the values of e/m (charge/mass) is

e, p, n, α
n, p, e, α
n, p, α, e
n, α, p, e

Solution

^e⁄_m for (i) neutron=⁰⁄₁=0

(ii) α-particle=²⁄₄=0.5

(iii) proton=¹⁄₁=1

(iv) electron=\(\frac{1}{1/1837}=1837\)

What is the ratio of mass of an electron to the mass of a proton?

1 : 2
1 : 1
1 : 1837
1 : 3

The compound in which cation is isoelectronic with anion is:

NaCl
CsF
NaI
K₂S

Solution

NaCl :No. of e^- in Na⁺= At. No. of Na^-1
= 11 – 1 = 10
No. e^-in Cl^-= At. No. of Cl + 1
= 17 + 1 = 18
CsF :No. of e^- in Cs⁺= 55 – 1 = 54
No. of e^- in F^-= 9+ 1 = 10
NaI :No. of e^-in Na⁺= 11 – 1 = 10
No. of e^- in I^-= 53 + 1 = 54
K₂S :No. of e^-in K⁺= 19 – 1 = 18
No. of e^- in S^2-= 16 + 2 = 18

The maximum number of electrons in p-orbital with n = 6; m =0 is

Solution

For p orbital with n= 6 and m= 0 indicates 6p_z orbital.It contains maximum of 2 electrons with spins opposite to each other.

Which one of the following is expected to have largest size?

F^-
2O^2-
Al³⁺
N^3-

Solution

F^- O^2- Al³⁺ N^3-
10 10 10 10
9 8 13 7

No. of e–
Nuclear charge
(Number of protons)
All the four given species are isoelectronic and size of isoelectronic species decreases with increase in nuclear charge. Among the four concerned atoms, N has lowest atomic number (nuclear charge), hence N^3-ion will be largest in size.

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